Question

An LC circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. The resistance of the circuit is negligible. At the instant, the circuit is closed be $t=0$. At what time is the energy stored completely magnetic

• A. $t=0ms$
• B. $t=1.54ms$
• C. $t=3.14ms$
• D. $t=6.28ms$

Answer 1

The correct option is B

$t=1.54ms$

Step 1: Given data.

The inductance of the inductor, $L=20mH$

The capacitance of the capacitor, $C=50\mu F$

Charge, $q=10mC$

Step 2: Calculating frequency

We know that,

The natural frequency of the LC circuit is given by,

$v=\frac{1}{2\pi \sqrt{LC}}$, where “v” is the frequency.

$\Rightarrow$$v=\frac{1}{2\times 3.14\times \sqrt{20\times {10}^{-3}\times 50\times {10}^{-6}}}$

$\Rightarrow$$v=159.24Hz$

Step 3: Calculating time period

Now,

Time period (T) ,

$T=\frac{1}{v}=\frac{1}{159.24}=6.28ms$

Since, we know that at time, $t=\frac{T}{4}$ the energy stored is completely magnetic.

Therefore,

$t=\frac{6.28}{4}$

$\Rightarrow$$t=1.57ms$

Hence, option B is correct. The time $t=1.57ms$ is nearly $t=1.54ms$.